In a transformer, where is current induced?

A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now

Vesnalui [34]

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere= $\rho$

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

$Density=\frac{mass}{volume}$

$Mass=Density\times volume=Constant$

Therefore, $\rho_1 V_1=\rho_2V_2$

Volume of sphere= $\frac{4}{3}\pi r^3$

Using the formula

$\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3$

$\rho R^3=\rho_2\times \frac{R^3}{8}$

$\rho_2=8\rho$

Hence, the density of the compressed sphere= $8\rho$

Option a is correct.

7 0

2 years ago

Which of the following will not have the same number of valence electrons?

nexus9112 [7]

Carbon is the answer to the problem

4 0

3 years ago

The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen

Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer: A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life = 5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 = 5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0

3 years ago

Which of the following happens to an object in uniform circular motion?

podryga [215]

Answer:

As an object moves in a circle, it is constantly changing its direction. ... Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed.

5 0

3 years ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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